Trying to build an 11x9 matrix from another 11x9 matrix and other elements

I am writing a program that compares different heat exchangers in order to select the best one for the application. I have successfully built an 11x9 array for the overall heat transfer coefficient (U), but now I am trying to build another 11x9 matrix for the log mean temperature difference (delTlm), and I am getting errors or answers that don't make sense. For example, when I did get an output for the delTlm matrix, every row had all values as the same. Here is the code that I have so far.

A = [2.4; 4.3; 7.4; 4.3; 7.4; 9.1; 2.4; 4.3; 7.4;].*0.0929; %m^2

%From problem 1, the exit temperature curves significantly flatten out
%after an NTU of 3 (returns diminish quickly). I will run the program from 
%2.5 NTU to 3.5 NTU
NTU = (2.5:0.1:3.5).';
for i = 1:length(NTU)
    
        %Calculating the effectiveness using the HE_effectiveness program 
        eff(i) = effectiveness(NTU(i), c, 'One Shell Pass');
    
        %Finding the actual heat transfer from Qmax and the effectiveness
        Q = (eff.*Qmax); %kW
    
        %Finding the overall transfer coefficient 
        U = (NTU.*Cmin)./A.'; %kW/m^2*deg C
        
        %Finding the log mean temperature difference
        deltaTlm = Q./U.*A.'
        
end

[KSSV]

The variable c is nor given,,,,the function effectiveness is not given..how you expect us to help you?

[Michael Costa]

%The HE is adiabatic.  
%Specific heat capacity of water at 60 deg C
C_c = 4.185*m4; %kJ/s*deg C

%The brine is modelled as pure water.
C_h = 4.185*m13; %kJ/s*deg C

%Solution
%---------
%Finding Cmin, Cmax, Qmax, and c
Cmin = min(C_c, C_h); %kJ/s*deg C
Cmax = max(C_c, C_h); %kJ/s*deg C
Qmax = Cmin*(T13 - T4); %kW
c = Cmin/Cmax;

[Michael Costa]

As I said in the question, I have successfully built the U matrix (thanks to another guy on here)...so you shouldn't have to worry about the c variable or the effectiveness function. Do you see either one of those in the calculation of the matrix that I am asking about? Maybe you have misunderstood my question.

Answer 1

[Mathieu NOE]

hello

there were only a few minor indexing issues in your main code.

also I recommend not using i or j as index var names as they are reserved for imaginary / complex numbers

so , here main code fixed :

%The HE is adiabatic.  
%Specific heat capacity of water at 60 deg C
C_c = 4.185*m4; %kJ/s*deg C
%The brine is modelled as pure water.
C_h = 4.185*m13; %kJ/s*deg C
%Solution
%---------
%Finding Cmin, Cmax, Qmax, and c
Cmin = min(C_c, C_h); %kJ/s*deg C
Cmax = max(C_c, C_h); %kJ/s*deg C
Qmax = Cmin*(T13 - T4); %kW
c = Cmin/Cmax;

A = [2.4; 4.3; 7.4; 4.3; 7.4; 9.1; 2.4; 4.3; 7.4;].*0.0929; %m^2
%From problem 1, the exit temperature curves significantly flatten out
%after an NTU of 3 (returns diminish quickly). I will run the program from 
%2.5 NTU to 3.5 NTU
NTU = (2.5:0.1:3.5)';

for ci = 1:length(NTU)
    
        %Calculating the effectiveness using the HE_effectiveness program 
        eff(ci) = effectiveness(NTU(ci), c, 'One Shell Pass');
    
        %Finding the actual heat transfer from Qmax and the effectiveness
        Q = (eff(ci).*Qmax); %kW
    
        %Finding the overall transfer coefficient 
        U = (NTU(ci).*Cmin)./A.'; %kW/m^2*deg C
        
        %Finding the log mean temperature difference
        deltaTlm(ci,:) = Q./U.*A';
        
end

[Michael Costa]

Awesome thank you Mathieu. This worked great.

[Mathieu NOE]

Glad it helped !